This document discusses two methods for obtaining partial fraction decompositions of rational functions. Method 1 writes the decomposition in a form that avoids complex numbers. Method 2 is preferable when there are distinct or repeated first-degree factors, as it eliminates the need to solve simultaneous equations. The document illustrates the two methods using examples and explains that experience will guide the appropriate choice of method for different situations.

1. Control Theory KHMak
1
Supplementary Material on Partial Fractions
Consider the following rational functions
1
1 2
1
( ) 1
( )
( ) ( 1)( 2 2)
N s s
F s
D s s s s
−
= =
+ + +
,
2
2 2
2
( ) 1
( )
( ) ( 1)( 2)
N s s
F s
D s s s
−
= =
+ +
.
We illustrate the use of two different (but equivalent) methods to obtain their partial fraction
decompositions.
Method 1
Typically, for functions like F1(s) which involve complex roots, we write the decomposition
as
1 2 2
1
( )
( 1)( 2 2) ( 1) ( 2 2)
s A Bs C
F s
s s s s s s
− +
= = +
+ + + + + +
.
This allows us to avoid the manipulation of complex numbers (which can be error-prone).
Multiplying both sides with the denominator D1(s) gives
2
1( ) 1 ( 2 2) ( )( 1)N s s A s s Bs C s= − = + + + + + .
Expanding and collecting powers, we have
2
1 ( ) (2 ) (2 )s s A B s A B C A C− = + + + + + + .
Comparing coefficients gives us 3 simultaneous equations which can be solved for the
unknown constants:
2A = − , 2B = , 3C =
Note that Method 1 is very general in that it can be used irrespective of the type of roots in
the denominator. However, when we have either distinct or repeated first-degree factors,
Method 2 is possibly a better choice since it eliminates the need to solve simultaneous
equations (of course, there is no reason why they cannot be used in combination)

2. Control Theory KHMak
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Method 2
F2(s) has one distinct and one repeated first-degree factor, so the decomposition has the form
2 2
( )
1 ( 2) 2
A B C
F s
s s s
= + +
+ + +
.
To find A, we multiply both sides by 1s + to cancel its denominator:
2 2 2
1
( ) ( 1) ( 1) ( 1)
( 2) ( 2) 2
s B C
F s s A s s
s s s
−
⋅ + = = + + + +
+ + +
.
Setting 1s = − eliminates the other terms and yields the value of A:
2 1 2
1
1
( ) ( 1) | 2
( 2)
s
s
s
A F s s
s
=−
=−
−
= ⋅ + = = −
+
.
B can be evaluated similarly by
2
2 2
2
1
( ) ( 2) | 3
( 1)
s
s
s
B F s s
s
=−
=−
−
= ⋅ + = =
+
.
C presents a problem since 2s = − is a pole of
2
1
( ) ( 2) ( 2)
( 1)( 2) ( 1) ( 2)
s A B
F s s s C
s s s s
−
⋅ + = = + + +
+ + + +
.
To avoid this, we proceed as with B
2 2
2
1
( ) ( 2) ( 2) ( 2)
( 1) ( 1)
s A
F s s s B C s
s s
−
⋅ + = = + + + +
+ +
,
but then differentiate once to “extract” C from the term Cs as well as eliminate B. Setting
2s = − then eliminates the first term and yields
2
2 2
2 2
1 1
( ) ( 2) 2
( 1) 1s s
d s
C F s s
ds s s=− =−
 −
 = ⋅ + = − + =   + + 
In summary, practice and experience will guide you in your choice of method (or
combination of methods) to use for partial fraction expansion.